Non-normal data

Where data does not meet the assumptions of parametric tests, we have two options:

1. Transform the data, and continue with parametric tests; or
2. Use non-parametric “equivalents” of parametric tests at the cost of power and loss of information.

Model-based inferential techniques

Resampling techniques

Randomisation or Bootstrap?

Why would these techniques work?

At the core of the resampling approach is the idea that the observed data is a random sample from a larger population.

Example: comparing two groups

Steps

Data

The sleep dataset in R contains the average extra hours of sleep, compared to control, for 10 patients who were given two different drugs.

library(tidyverse)
glimpse(sleep)

Rows: 20
Columns: 3
$ extra <dbl> 0.7, -1.6, -0.2, -1.2, -0.1, 3.4, 3.7, 0.8, 0.0, 2.0, 1.9, 0.8, …
$ group <fct> 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
$ ID    <fct> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Are assumptions of normality met?

We picked a dataset where the assumptions are met, so that we can compare the results with the parametric test.

ggplot(sleep, aes(x = group, y = extra)) +
  geom_boxplot()

Calculating the t-test statistic

Recall that the test statistic for the two-sample t-test is:

t = \frac{Difference\ in\ the\ means}{Standard\ error\ of\ the\ difference}

We could calculate it manually, but let’s just use the t.test() function in R since the function calculates the test statistic for us. For example, the observed test statistic for the sleep data is:

t.test(extra ~ group, data = sleep)$statistic

        t 
-1.860813 

Step 1: Pool the scores

The first step is to pool the data. The pooled data is:

pooled_data <- sleep$extra
pooled_data

 [1]  0.7 -1.6 -0.2 -1.2 -0.1  3.4  3.7  0.8  0.0  2.0  1.9  0.8  1.1  0.1 -0.1
[16]  4.4  5.5  1.6  4.6  3.4

Where the first 10 scores are from the first group, and the next 10 scores are from the second group.

Step 2: Randomly reassign the scores

pooled_data

 [1]  0.7 -1.6 -0.2 -1.2 -0.1  3.4  3.7  0.8  0.0  2.0  1.9  0.8  1.1  0.1 -0.1
[16]  4.4  5.5  1.6  4.6  3.4

Next, we randomly shuffle the pooled data and re-assign the first 10 scores to group 1, and the next 10 scores to group 2.

set.seed(1022)
shuffled_data <- sample(pooled_data)
shuffled_data

 [1] -1.2  2.0  3.7  0.1  0.8  1.6  0.8  1.9  0.0  3.4 -1.6 -0.2  1.1 -0.1  0.7
[16]  4.4  5.5  3.4 -0.1  4.6

group1 <- shuffled_data[1:10]
group2 <- shuffled_data[11:20]

Step 3: Calculate the test statistic

We’re not using the results from the t.test() function, but just extracting the test statistic.

t.test(group1, group2)$statistic

         t 
-0.4995608 

Step 4: Repeat many times

Putting it all together, we can write a function to obtain the test statistic:

random_t <- function(data) {
  shuffled_data <- sample(data)
  group1 <- shuffled_data[1:10]
  group2 <- shuffled_data[11:20]

  return(t.test(group1, group2)$statistic)
}

Repeat the function 10,000 times:

set.seed(1034)
random_t_values <- replicate(10000, random_t(pooled_data))

Step 5: Compare the observed test statistic

Finally, we can compare the observed test statistic to the randomised distribution. This can be done by calculating the proportion of randomised test statistics that are more extreme than the observed test statistic.

fit <- t.test(extra ~ group, data = sleep) # test on observed data
p_value <- mean(abs(random_t_values) >= abs(fit$statistic))
round(p_value, 2)

[1] 0.08

How does this compare to the parametric t-test?

If we round the p-values of both tests to two decimal places, we get:

round(fit$p.value, 2)

[1] 0.08

As we can see, the p-values are very similar. This is because the assumptions of the parametric test are met, so the results will be close even though the methods are different!

What’s the difference between the two techniques?

Visualising the randomised distribution

ggplot(data.frame(t = random_t_values), aes(x = t)) +
  geom_histogram(binwidth = 0.5, fill = "skyblue", color = "black") +
  geom_vline(xintercept = fit$statistic, color = "red", linetype = "dashed") +
  labs(title = "Randomised distribution of t-test statistic",
       x = "t-test statistic",
       y = "Frequency")

We can see that the observed test statistic is well within the distribution of the randomised test statistics (which is normally distributed).

Randomisation tests using coin

The coin package in R provides a simple interface to perform randomisation tests.

library(coin)

Let’s use the sleep dataset to demonstrate how to use the coin package to perform a randomisation test.

perm_test <- independence_test(extra ~ group, data = sleep, distribution = approximate(B = 9999))

# Print results
print(perm_test)

    Approximative General Independence Test

data:  extra by group (1, 2)
Z = -1.7508, p-value = 0.08321
alternative hypothesis: two.sided

Example: estimating the mean

Steps

Data

The BOD dataset in R contains the biochemical oxygen demand (mg/L) measurements of 6 samples over time.

glimpse(BOD)

Rows: 6
Columns: 2
$ Time   <dbl> 1, 2, 3, 4, 5, 7
$ demand <dbl> 8.3, 10.3, 19.0, 16.0, 15.6, 19.8

Step 1: Resample the data

From the pooled original data:

BOD$demand

[1]  8.3 10.3 19.0 16.0 15.6 19.8

We sample() with replacement:

set.seed(1113)
resampled_data <- sample(BOD$demand, replace = TRUE)
resampled_data

[1] 15.6  8.3  8.3  8.3 16.0  8.3

Noting that some scores will be repeated, and some will be missing.

Step 2: Calculate the parameter of interest

The parameter of interest is the mean value.

mean(resampled_data)

[1] 10.8

Step 3: Repeat many times

Since it’s a simple process, we can write a function to calculate the mean:

bootstrap_mean <- function(data) {
  resampled_data <- sample(data, replace = TRUE)
  return(mean(resampled_data))
}

Then repeat the function 10,000 times:

set.seed(1116)
bootstrap_means <- replicate(10000, bootstrap_mean(BOD$demand))

Step 4: Calculate the 95% CI

The 95% CI is calculated from the simulated distribution:

meanval <- mean(bootstrap_means)
CI <- quantile(bootstrap_means, c(0.025, 0.975))

Putting it together, the mean is 14.85131 with a 95% CI of [11.25, 18.13].

How does this compare to the parametric test?

If we use the t.test() function to calculate the 95% CI:

t.test(BOD$demand)

    One Sample t-test

data:  BOD$demand
t = 7.8465, df = 5, p-value = 0.0005397
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
  9.973793 19.692874
sample estimates:
mean of x 
 14.83333 

How different are the results?

Again we can use a package to do this

The boot package in R provides a simple interface to perform bootstrap tests.

library(boot)

Let’s use the BOD dataset to demonstrate how to use the boot package to perform a bootstrap test.

# Define function to compute mean
boot_mean <- function(data, indices) {
  return(mean(data[indices]))
}

boot_test <- boot(BOD$demand, boot_mean, R = 10000)

# Print results 
print(boot_test)

ORDINARY NONPARAMETRIC BOOTSTRAP


Call:
boot(data = BOD$demand, statistic = boot_mean, R = 10000)


Bootstrap Statistics :
    original    bias    std. error
t1* 14.83333 -0.010455     1.72223

# Compute bootstrap confidence interval
boot.ci(boot_test, type = c("perc"))

BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
Based on 10000 bootstrap replicates

CALL : 
boot.ci(boot.out = boot_test, type = c("perc"))

Intervals : 
Level     Percentile     
95%   (11.40, 18.13 )  
Calculations and Intervals on Original Scale

Using infer

library(tidymodels)

Let’s use the sleep dataset to demonstrate how to use the infer package to perform a randomisation test (also makes it easier to compare against manual method).

The infer package requires the user to use an expressive grammar to specify the analysis.

Steps

1. specify() the response variable of interest, then
2. hypothesise() the null hypothesis, then
3. generate() the null distribution, and finally
4. calculate() the p-value.

Two-sample t-test using infer

First we need to calculate the observed test statistic so that we can compare it to the simulated distribution.

observed <- sleep %>%
  specify(extra ~ group) %>%
  calculate(stat = "diff in means", order = c(1, 2))
observed

Response: extra (numeric)
Explanatory: group (factor)
# A tibble: 1 × 1
   stat
  <dbl>
1 -1.58

Then we generate the null distribution and calculate the p-value:

set.seed(1034)
pval_infer <- 
  sleep %>%
  specify(extra ~ group) %>%
  hypothesise(null = "independence") %>%
  generate(reps = 10000, type = "permute") %>%
  calculate(stat = "diff in means", order = c(1, 2)) %>%
   get_p_value(obs_stat = observed,
              direction = "two-sided")

What are the differences?

As we can see, the results are very similar because the assumptions of the parametric test were already met.

Example: beetles

The beetle dataset was used in last week’s lecture to demonstrate the non-parametric Wilcoxon rank-sum test.

beetle <- readr::read_csv("data/beetle.csv")
glimpse(beetle)

Rows: 45
Columns: 2
$ SIZE    <chr> "small", "small", "small", "small", "small", "small", "small",…
$ BEETLES <dbl> 256, 209, 0, 0, 0, 44, 49, 117, 6, 0, 0, 75, 34, 13, 0, 90, 0,…

Assumption

Recall that the data does not meet the assumptions of normality:

p1 <- ggplot(beetle, aes(x = BEETLES)) +
  geom_histogram(bins = 14, position = "dodge") +
  facet_wrap(~SIZE, ncol = 1)
p2 <- ggplot(beetle, aes(x = SIZE, y = BEETLES)) +
  geom_boxplot()
p3 <- ggplot(beetle, aes(sample = BEETLES)) +
  stat_qq() +
  stat_qq_line() +
  facet_wrap(~SIZE, ncol = 1)

library(patchwork)
p1 + p2 + p3

T-test via resampling using infer

First, calculate the test statistic:

observed <- beetle %>%
  specify(BEETLES ~ SIZE) %>%
  calculate(stat = "diff in means", order = c("small", "large"))

Then generate the null distribution and calculate the p-value:

set.seed(1034)
pval_infer <- 
  beetle %>%
  specify(BEETLES ~ SIZE) %>%
  hypothesise(null = "independence") %>%
  generate(reps = 10000, type = "permute") %>%
  calculate(stat = "diff in means", order = c("small", "large")) %>%
  get_p_value(obs_stat = observed,
              direction = "two-sided")

pval_infer

# A tibble: 1 × 1
  p_value
    <dbl>
1   0.025

Comparisons

Let’s compare the p-values from

1. a t-test (if we ignore violations of assumptions),
2. the wilcoxon rank-sum test, and
3. the infer randomisation test.

Comparisons

T-test

p_ttest <- t.test(BEETLES ~ SIZE, data = beetle)$p.value %>%
  round(3)

Wilcoxon rank-sum test

p_wilcox <- wilcox.test(BEETLES ~ SIZE, data = beetle)$p.value %>%
  round(3)

Randomisation test

p_infer <- pull(pval_infer, p_value) %>%
  round(3)

Results

• As we can see, the p-values are quite different because the assumptions of the parametric test were violated.
• The randomisation test is more robust and provides a more accurate estimate of the p-value than the Wilcoxon rank-sum test.

How to report results of randomisation test

The results of the randomisation test can be reported as follows:

Beetle consumption was significantly different between small and large beetles (t = 2.19, R = 10000, p = 0.025).